Min Distance Calculations 2

[pmarc]

If you don't mind, I would like to hear some answers to the question from the attachment.
Please try to imagine really the worst-case than can ever happen for this type of part.
Drawing is in accordance with ASME Y14.5-2009.

http://files.engineering.com/getfil...-487f-92f3-159e42fe7896&file=Min_distance.png

 

[SeasonLee]

pmarc
Will the minimum distance be different or keeps at 4.7 if the inner hole with a little bow (I/O straight) on your sketch?

Season

 

[greenimi]

Thank you Pmarc,
Interesting stuff.
I remember that not too long ago, on a similar thread about minimum distance calculation you said that the form error DOES NOT need to be taken in consideration for the stackup (Min distance X). I would say the instances are ”almost” similar. In the thread below (Kurlikovski Fundamentals book thread, Fig. 9.14) we start the calculations from 69.6 (again with no form error considered)

http://www.eng-tips.com/viewthread.cfm?qid=348558

I am trying to find a reason why the calculations now are different and one thing that comes in my mind is that in your case (this thread) primary datum is A (the one which gave you the form error), but in the “Kurlikovski thread” the form error is not considered because datum B is secondary, or maybe better said because 70.4/69.6 width/length is NOT the primary datum?
Otherwise I would see both situations, again almost similar (if I take your solution sketch, the one with the bow shown, and turn the page 90° degree counterclockwise, replace 20.1/19.9 dimension with 70.4/69.6 AND hole size 10.1/9.9 to 8.6/8.0 AND positional callout Ø 0(MMC) with Ø1 (MMC), I don’t see any other reason on why the form error is available in one case but not the other).
Are you with me? Do you understand what I am talking about?
No, I am not saying you are wrong, I am just trying to learn from both examples.

 

[pmarc]

Season,
On my sketch the "little bow" of ID resulting in potential reduction of mininimum 4.7 can't happen. Dia. 10.1 is the size of perfect envelope, and for this size of the envelope presence of such bow would mean that requirement for actual local size to be smaller than or equal to 10.1 has been violated.

You can also think about it in terms of outer boundary of the hole (that is, virtual condition boundary for case #1, resultant condition boundary for case #2), centered perfectly at datum axis A. Within this boundary material of the hole must always be. In both cases the size of outer boundary is dia. 10.3. That means the distance from datum axis A to one side of the outer boundary (to the bottom side on my picture) is 10.3/2 = 5.15. Now, if you look again at my picture, you will notice that the very same happens there. The distance from datum axis A to the bottom edge of the hole is 0.1 + 10.1/2 = 0.1 + 5.05 = 5.15.

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greenimi,
There is nothing wrong in saying that someone is wrong. Everyone makes mistakes and I am no exception. It is really great thing that you ask questions and try to learn as much as possible. I also learn a lot when I try to prove my standpoint, believe me.

I understand what you are talking about, however I do not agree that these two examples are "almost" similar (and it has nothing to do with datum features being primary or secondary).

The two examples would be geometrically identical IF in my example datum was derived from upper "edge" of the datum feature simulator and not from its diameter. Try to imagine this for a moment. You will notice that the minimum possible distance from that datum to the closest portion of opposite side of the cylinder is 19.9 in my case, and can't be smaller. In your case this would be 69.6 and nothing less.

 

[greenimi]

Pmarc,
Okay, I understood.
And to expand a little bit: would you agree that IF in the Kurlikovski thread the primary datum feature were 70.4/69.6 width (driving the primary datum to be the middle plane) then the minimum distance calculation should take in consideration the form error: 70.4-69.6 = 0.8?
Thank you

 

[pmarc]

Yes, I would agree with that.

 

[SeasonLee]

Thanks for your feedback, pmarc. Its an interesting thread.

Season

 

[pmarc]

No problem. I am really glad that it turned out to be interesting lesson for you.

 

[gabimo]

Pmarc,
I am trying to learn stack-up tolerances and I went over your exercise and –if you don’t mind –I have a couple of questions regarding your proposed solution:

1.) Is it true that when a geometric tolerance with Least material Condition is specified than the rule #1 is not applicable? In other words among the exceptions to rule#1 is a geometric tolerance with LMC specified.

2.) How the minimum in your cases #1 and #2 would look like IF datum feature is switched from OD to ID?
ID: Ø10.1/9.9 datum feature A and
OD Ø20.1/19.9 with position:
Case#1 position Ø0.2 at LMC to A
Case#2 position Ø 0 at MMC to A

Would the min distance be the same as your original posting? (was 4.7 in both cases)

Thank you

 

[pmarc]

gabimot,
1. It is NOT TRUE that when a geometric tolerance with Least Material Condition is specified, the rule #1 is not applicable. When it happens perfect form at LMC is required. This is not directly stated in para. 2.7 of Y14.5M-1994 where Rule #1 is described, but is expressed in second sentence of para. 5.3.5 (page 89). In Y14.5-2009 this requirement is directly given in para. 2.7(d).

2. I get the same result, that is 4.7, in both cases when the features are switched.

 

[gabimo]

pmarc,
I am lost on this:
"Quote:

[Due to the possible bow of the cylinder = 0.2, the 19.9 diameter is not centered at datum axis A]"
12 Sep 13 8:27

What is datum A? Isn't it the minimum inscribed perfect cylinder you can fit around the OD? How can NOT be centered at datum axis A since this cylinder defines datum axis A?

Now, since we are interested in finding the minimum distance that implies 19.9 OD to have a perfect form so, why are we substracting another 0.2? Again, let me be clear, I am trying to understand and I am not saying you are wrong on this.

If I am using virtual boundary/resultand boundary I got: 19.9-10.3 = 4.8
Your original case#1: pos Ø.02L with A: Virtual boundary/condition: 10.3 = material free zone.
Your original case#2: pos Ø.0M with A: Resultant boundary/condition: 10.3 = material containing zone.

What am I missing? If we are interested in finding the minimum (extreme) because of rule#1 any form error would have served to increase our minimum wal thickness.

 

[gabimo]

I should have said:
What is datum A? Isn't it the AXIS of the minimum inscribed perfect cylinder you can fit around the OD?

 

[pmarc]

gabimot,
Since I do not have access to editable version of my original graphic at the moment, words will have to suffice.

1. In my original scheme datum axis is the axis of perfect minimum CIRCUMSCRIBED cylinder that can be contracted about surface of the OD. It is obvious that the actual mating envelope of the OD must be centered at datum axis, because the axis is derived from that envelope. But actual local size of the OD in the middle of its length (19.9 = LMC on my graphic) does not have to be centered at the datum axis - the graphic clearly shows that.

2. Here is the trap with virtual boundary/resultant boundary approach:
In my original graphic, it is true that the outer boundary of the ID is 10.3 (10.1+0.2) for cases #1 and #2, however...
It is not true that the size of the inner boundary of the OD is 19.9.
The inner boundary of the OD is 19.9 MINUS maximum allowable form (straigthness) error. Since total size tolerance for the OD is 20.1-19.9 = 0.2, the maximum allowable form error of the OD is exaclty 0.2 (because of Rule #1). So that gives you 19.9-0.2 = 19.7 size of the inner boundary of the OD.
Now, (19.7-10.3)/2 = 9.4/2 = 4.7.

If that still does not convince you, take a look at fig. 2-6 in Y14.5M-1994 (upper pictures showing the shaft). Notice that the dia. 20 (LMC) dimension in the middle of shaft's height is also not centered at the axis of what they called MMC Perfect Form Boundary. And if on the very same figure you draw a cylinder within which there is always a material of the shaft, its size will not be 20.0, but 19.9.

Does it help?

 

[gabimo]

Pmarc,

Now, I do understand perfectly what is your approach to this small stack-up issue (minimum wall thickness). Thank you very much.


I have a stack-up tolerance book and an example in this book confuse me more than it probably should.

Its saying that the form error has not been included intentionally in the calculations because:

for the MMC specification, because of rule#1 and because we are interested in the extremes, any form error would have served to increase the minimum wall thickness or to decrease the maximum wall thickness.

for LMC specification is then when we specify LMC in the feature control frame, perfect form at LMC is required. That is why we specify LMC in the first place-to maintain a minimum wall thickness. The end result is that the LMC calculations are valid, because any form error would have to have gone away from our minimum wall.

Just worth to mention that in the book I am talking about the OD and ID datum features are switched versus your example (in my book the primary datum feature is the ID)

 

[pmarc]

gabimot,
I do not know what book you have. Maybe in the example from this book the meaning of "minimum wall thickness" is different than in my case. Hard to tell without seeing it.

I just want to repeat that the difference between 4.7 and 4.8 in my example does not come from what is happening with toleranced feature (either controlled by position at MMC or LMC). It comes from proper understanding of size of inner boundary of datum feature OD. And it will come from proper understanding of size of the outer boundary of the ID, if the ID has been chosen as datum feature.